INTRODUCTION:

Increased
access to transport is a feature of today’s society. Most people access some
form of transport for travel to and from school or work and for leisure outings
at weekends or on holidays. When describing journeys that they may have taken in
buses or trains, they usually do so in terms of time or their starting point and
their destination. When describing trips they may have taken in planes or cars,
they normally use the time it takes, distance covered or the speed of the
vehicle as their reference points. While distance, time and speed are fundamental to
the understanding of kinematics and dynamics, very few people consider a trip in
terms of energy, force or the momentum associated with the vehicle, even at low or
moderate speeds.

The faster a vehicle is travelling, the further it will go
before it is able to stop. Major damage can be done to other vehicles and to the
human body in collisions, even at low speeds. This is because during a collision some or all
of the vehicle’s kinetic energy is dissipated through the vehicle and the
object with which it collides. Further, the materials from which vehicles are
constructed do not deform or bend as easily as the human body. Technological
advances and systematic study of vehicle crashes have increased understanding of
the interactions involved, the potential resultant damage and possible ways of
reducing the effects of collisions. There are many safety devices now installed
in or on vehicles, including seat belts and air bags. Modern road design takes
into account ways in which vehicles can be forced to reduce their speed.

The branch of Physics that deals with phenomena such as
those mentioned above is called mechanics.

Note: Some internet browsers (eg Firefox) do not
accurately display text symbols such as Greek letters used to represent
quantities in Physics. For example, capital delta is displayed as D in
Firefox. So, if you see below something like Ds/Dt for speed or velocity,
it is meant to appear as delta s/delta t. This is just something to be
aware of in case you do come across such issues. The square root sign is
another one not displayed properly by some browsers. Any symbols used in
equations produced by equation editors will of course display properly.

MECHANICS:

The branch of Physics that is concerned with the motion and
equilibrium of bodies in a particular frame of reference is called “mechanics”.Mechanics can be divided into three branches: (i) Statics –
which deals with bodies at rest relative to some given frame of reference, with
the forces between them and with the equilibrium of the system; (ii) Kinematics
- the description of the motion of bodies without reference to mass or force;
and (iii) Dynamics – which deals with forces that change or produce the
motions of bodies.

Some common terms used in the study of mechanics (and
indeed many other branches of Physics) are: scalars, vectors and SI Units.

2.Vectors – A vector is a physical quantity defined in terms of both
magnitude and direction - eg force, velocity, acceleration, electric field
strength

. Diagramatically we can represent a vector by a straight line with an
arrow on one end.The length of the
line represents the magnitude of the vector quantity and the direction in which
the arrow is pointing represents the direction of the vector quantity.We will say much more about vectors later in this topic.

3.Systeme International (SI) Units – The internationally agreed
system of units.There are seven
fundamental units.The three that
we will use in this topic are the metre (length), the kilogram (mass) and the
second (time).Various prefixes
are used to help express the size of quantities eg a nanometre (1 nm) = 10-9
of a metre, a gigametre (1 Gm) = 109 metres. See Appendix C,
p.243 of Excel Preliminary Physics by Warren for a list of some common SI units
and prefixes. Most texts will contain such information.

Since describing the features of something is usually a lot
simpler than explaining how or why it works, we will start with a look at
kinematics.

MOTION:

1.Displacement – is the distance of a body from a given point in a
given direction.It is a vector
quantity.The SI unit of
displacement is the metre (m).

2.Speed – The speed of a body is the rate at which it is covering
distance.It is a scalar quantity.The SI units are m/s, which can also be written as ms-1.

where vav
= average speed, d = total distance travelled and t = total time taken to travel
distance d.

3.Velocity – The velocity of a body is its speed in a given
direction.In other words, velocity
is the rate of change of displacement with time.It is a vector quantity with the same SI units as speed.

where vav =
average velocity, Dr = change in
displacement and Dt = change in
time taken to achieve that change in displacement.

Another way to express average velocity is as the average of the initial and
final velocities.

where vav
= average velocity, u = initial velocity of the body and v = final velocity of
the body.Note that this
equation applies ONLY when the velocity of the body is increasing or decreasing
at a constant rate.

4.Acceleration – The acceleration of a body is the rate of change of
the velocity of the body with time.It is a vector quantity, with units of
(metres/second)/second, written as ms-2.

where aav = average acceleration, Dv
= change in velocity of the body and Dt
= change in time over which the change in velocity took place.Alternatively, we may write:

where aav = average acceleration,
v = final velocity of the body, u =
initial velocity of the body, and t = time over which the change in velocity
took place.

MOTION GRAPHS:

Often the most effective way to describe the motion of a
body is to graph it. Note that in this section the variable “s”
will be used to represent displacement instead of “r”. You
will find in Physics that these two variables are both in common use to denote
displacement.

Displacement-Time Graphs

These may be used to gain information about the
displacement of an object at various times or about the velocity of the object
at various times.

Clearly, the gradient (slope) of a displacement-time graph
gives the velocity.

Gradient = DS/Dt= velocity

Note that a positive gradient implies a positive velocity
and a negative gradient implies a negative velocity.

For a curved displacement-time graph, the gradient of the tangent to the
curve at a particular point equals the gradient of the curve at that point,
which in turn equals the velocity of the object at that particular time.Such a velocity, that is, the velocity at a particular instant in
time, is called the instantaneous velocity.

An example of an instrument that measures instantaneous
velocity is the speedometer in a car.In
older cars the speedometer was linked mechanically to the
transmission. These days, however, a device located in the transmission produces
a series of electrical pulses whose frequency varies in proportion to the
vehicle's speed.The electrical
pulses are sent to a calibrated device that translates the pulses into the speed
of the car. This information is sent to a device that displays the vehicle's
speed to the driver in the form of a deflected speedometer needle or a digital
readout.

Note that a straight line displacement-time
graph implies that velocity is constant.A
curved line displacement-time graph implies that velocity is changing with time
(ie the object is accelerating).

Velocity-Time Graphs

These may be used to gain information about the
displacement, velocity and acceleration of an object at various times.

The gradient is clearly the acceleration of the object.

Gradient = Dv/Dt= acceleration

Note that a positive gradient implies a positive
acceleration and a negative gradient implies a negative acceleration.

Also, the area under the graph,

in the case above, has units of: seconds x metres per
second = metres.Thus, the area
under a velocity-time graph is equal to the displacement travelled by the object
in the time Dt.

A non-horizontal, straight line velocity-time graph implies
that acceleration is constant and non-zero.

A curved line velocity-time graph implies that acceleration
is varying.

Acceleration-Time Graphs

These may be used to gain information about the velocity
and acceleration of an object at various times.

The area under an acceleration-time graph gives the
change in
velocity of an object during the time interval Dt.Check the
units of the area: (ms-2 x s = ms-1).

A horizontal straight line acceleration-time graph implies
that velocity is varying at a constant rate (ie velocity is increasing or
decreasing by the same amount each second).That is, acceleration is constant.

RELATIVE VELOCITY:

Often it is necessary to compare the velocity of one object
to that of another.For instance,
two racing car drivers, A and B, may be travelling north at 150 km/h and 160
km/h respectively.We could say
that the velocity of car B relative to car A is 10 km/h north.In other words, driver A would see driver B pull away from
her with a velocity of 10 km/h north.

Likewise, two jet aircraft, C and D, flying directly at
each other in opposite directions (hopefully as part of an aerobatics display)
may have velocities of 900 km/h north and 1000 km/h south respectively.We could say that the velocity of D relative to C is 1900 km/h south.In other words, jet C will observe jet D flying towards it at a speed of
1900 km/h.

Clearly, when the objects are travelling in the same
direction, the velocity of one relative to the other is the difference between
their speeds, taking due care to state the appropriate direction.When the objects are travelling in opposite directions the velocity of
one relative to the other is the sum of their speeds, again taking due care to
state the correct direction.There
is a vector equation which can be used to calculate the relative
velocities of objects, even when the objects travel at various angles to one
another but this equation is outside the scope of the present syllabus (for some
unfathomable reason).

FORCE:

What is Force?

In simple terms a force can be defined as a push or a pull.We experience examples of forces every day.If we push a stationary lawn mower (with enough force) it begins to move
– that is, it accelerates and its velocity increases.If we push on the brakes of a moving bicycle, it slows down – that is
it decelerates (or undergoes negative acceleration).If we apply sufficient force to an aluminum can, by squeezing it with
our hand, we can change the shape of the can.

So a force can cause a change in the state of motion of
an object or a change in the shape of an object.In fact, all accelerations (and decelerations) are caused by forces.

Does every force cause acceleration?

Again, from our everyday experience, we know the answer to
this question is “no”. If a person pushes on the brick wall of a house, the
house does not accelerate.Sometimes
when we want to push or pull an object from one place to another we find that no
matter how hard we push or pull, we just cannot move (accelerate) the object.

In previous Science courses, a qualitative
relationship was established between force, mass and acceleration.In the Preliminary Physics Course we need to establish a quantitative
relationship between these three quantities.

What is the relationship between force and
acceleration?

We could perform an experiment to determine the
relationship between the size of a force applied to an object at rest on a
laboratory bench and the change in velocity experienced by the object over a set
period of time (ie the acceleration).Such
an experiment would produce results as shown below.

The graph above shows that:

¨The change in velocity does not happen instantaneously. A certain
amount of force is required before the object begins to accelerate.This makes sense, since the force of friction between the
bench and the object must be overcome before the object can move.So, we can say that a net external force is required in order to
change the velocity of an object.

¨The acceleration produced is directly proportional to the force
applied.If we repeated the
experiment on a frictionless surface (eg using a dry-ice puck on a very smooth,
polished table top) the straight-line graph would even pass through the origin.

What is the relationship between acceleration and
mass?

We could measure the accelerations produced when the same
sized force is applied to different objects.Such an experiment would produce results like those below.

The graph above suggests that there is an inverse
relationship between acceleration and mass.A plot of acceleration versus the reciprocal of mass, using the same
data, would produce a graph similar to that below.

This graph clearly shows that:

¨The acceleration produced by a given force is inversely
proportional to the mass of the object.

NEWTON’S
LAWS OF MOTION

Newton’s First and Second
Laws:

By combining the results above and
defining the units of force appropriately, we can write that:

.

This can be taken as a
statement of Newton’s Second Law.The
SI Unit of force is the newton (N), defined so that 1N = 1kgms-2.

Note that in the above equation, F
is the vector sum of all the forces acting on the object, m is the mass
of the object and a is its vector acceleration.To remind us of that fact we will write:

.

Note that if the resultant force
on the object is zero, there is no acceleration.Therefore, in the absence of a resultant force, an object’s velocity
will remain unchanged.In other
words, an object at rest will remain at rest, and an object in motion will
remain in motion with uniform velocity, unless acted upon by a net external
force.This is a statement of
Newton’s First Law, which in fact is contained in the Second Law as a special
case (for SF = 0).

Newton’s Third Law:

Forces acting on a body originate
in other bodies that make up its environment.Any single force is only one aspect of a mutual interaction between two
bodies.We find by experiment that
when one body exerts a force on a second body, the second body always exerts a
force on the first.Furthermore, we
find that these forces are equal in size but opposite in direction.A single, isolated force is therefore an impossibility.

If one of the two forces involved
in the interaction between two bodies is called an action force, the
other is called the reaction force.Either force may be called the action and the other the reaction.Cause and effect is not implied here, but a mutual simultaneous
interaction is implied.

This property of forces was first
stated by Newton in his Third Law: “To every action there is always opposed
an equal reaction; or, the mutual actions of two bodies upon each other are
always equal, and directed to contrary parts.”

In other words, if body A exerts a
force on body B, body B exerts an equal but oppositely directed force on body A;
and furthermore the forces lie along the line joining the bodies.Notice that the action and reaction forces, which always occur in
pairs, act on different bodies.If
they were to act on the same body, we could never have accelerated motion
because the resultant force on every body would be zero.

Consider the following examples:

1.Imagine a boy kicking open a door.The force exerted by the boy B on the door D accelerates the door (it
flies open); at the same time, the door D exerts an equal but opposite force on
the boy, which decelerates the boy (his foot loses forward velocity).The force of the boy on the door and the force of the door on the boy is
an action-reaction pair of forces.

2.When you walk, you apply a force backwards on the earth.Likewise, the earth applies a force to you of equal magnitude but in the
opposite direction.So, you move
forwards.

The force of the person on the earth and the force of the earth on the person is
an action-reaction pair of forces.

3.Consider a body at rest on a horizontal table:

Each of the pairs of forces above is an action-reaction pair of forces.

USEFULNESS
OF VECTOR DIAGRAMS:

Many of the quantities with which we deal in Physics
are vectors.Sometimes
we need to add a number of vectors together.For instance, we may be trying to calculate the total or resultant force
acting on a car when several forces act on the car simultaneously – the wind,
friction, gravity and the force supplied by the engine.Sometimes we need to subtract two vectors.For instance, we may be trying to calculate the change in
velocity of a car as it goes around a bend in the road.The change in velocity of the car equals the final velocity of the car
minus initial velocity of the car.

When the need arises to add or subtract vector
quantities, this proves to be easy only when the vector quantities act along the
same straight line.If the vectors
act at an angle to each other we really need to draw a vector diagram to assist
in solving the problem.

Vector analysis is an extremely
important aspect of Physics and there are several different methods available to
add, subtract and even multiply vectors.Unfortunately,
the current Syllabus requires that you have only a very basic understanding of
vector analysis.So, we will
examine a single, simple but very useful method of adding and subtracting
vectors.

VECTOR ADDITION:

The method we will use is called the “Vector
Polygon” method.To find the
sum of a number of vectors draw each vector in the sum, one at a time, in the
appropriate direction, placing the tail of the second vector so that it just
touches the head of the first.Continue
in this fashion until all of the vectors in the sum have been included in the
diagram.Note that it does not
matter which vector you start with.

The vector that closes the vector polygon in the same sense
as the component vectors is called the equilibrant.It is the vector which when drawn into the diagram gets you back to where
you started.The vector that
closes the vector polygon in the opposite sense to the component vectors is
called the resultant.The resultant is the answer to the sum of all the vectors.Its size can be calculated mathematically or measured using a ruler if
the vector polygon has been drawn to scale.The direction of the resultant can be calculated mathematically or can be
measured using a protractor if the vector polygon has been drawn to scale.Either way, the direction of the resultant must be stated in an
unambiguous way.

Sometimes in Physics our vector additions only involve two
vectors at a time.In this case,
the polygon formed is a triangle, making the mathematical calculation of the
magnitude and direction of the resultant quite straight forward.So, keep your wits about you and bring your
knowledge of triangle geometry and trigonometry into the Physics classroom.

EXAMPLE
1:A fighter pilot flies her F-14D Tomcat jet with a true
airspeed of 400 km/h
North.A
crosswind from the East blows at 300 km/h relative to the ground.Calculate the jet’s resultant velocity relative to the ground.

Note:
For aircraft, the true airspeed (TAS) is the actual speed of the aircraft
through the air (the speed of the aircraft relative to the air). The wind speed
is usually measured relative to the ground. Groundspeed is the speed of the
aircraft relative to the ground. The groundspeed of the aircraft is the
vector sum of the true airspeed and the wind speed.

Obviously, a vector diagram would be very useful in solving
Example Problem 1.See the diagram
below.

By Pythagoras’ Theorem, the magnitude of
the resultant velocity of the jet is:

and the direction can be found using basic trigonometry
as follows:

So, the velocity of the jet relative to the ground is
500 km/h N36.9oW.

Note that if the angle between the two vectors being added
together is other than 90o, Cosine Rule and Sine Rule
can be used to solve the problem mathematically.Note also the use of the compass in the diagram to establish
direction.

EXAMPLE
2: In the previous problem, in which direction should the pilot
head and with what airspeed in order to actually fly north at 400 km/h relative to
the ground?

Again a vector diagram is useful.Our intuition tells us that the pilot must fly into the wind.So, when we draw a diagram that shows all of the information that we know
to be true, we obtain the diagram shown below.

Clearly, if the pilot flies N36.9oE with an
airspeed of 500 km/h, the wind will bring her back to a heading of due North at a
speed of 400 km/h relative to the ground.Remember
also, there is usually more than one way to give the direction.The direction the pilot should fly in this example could just
as correctly be given as E53.1oN or as a True Bearing of 36.9o.

EXAMPLE
3:Four children pull on a small tree stump firmly stuck in
the ground.Looking down on the
tree stump from above, the forces applied by the children are as shown below.

Determine the resultant force applied to the tree stump.

To solve this problem mathematically we would need to add
two of the vectors together, then add our answer to the third vector and finally
add our answer to that addition to the fourth vector.It is actually far quicker and easier to solve this problem graphically.To do this we construct a vector polygon, using the rules stated above
and simply measure the size and direction of the resultant.See the diagram below.

The resultant force, R, is found by measurement to be 3.1 N at an angle of
39o clockwise from the direction of the 4.5 N force.

Note that when using a graphical approach, the scale
must be clearly stated on the diagram.Always choose a sensible numerical scale.Also, choose a scale that will produce a large diagram.The larger the diagram, the more accurate the answer.For the example problem above, the scale used was 1 N = 1.5 cm.This is certainly the smallest scale I would use for this particular
problem.Anything less is too
inaccurate.A scale of 1 N = 2 cm
would be preferable.The smaller
scale was used here to fit the diagram neatly onto this page.

Note
also, that depending on which printer is used to print these notes, there may be
a small discrepancy between the stated scale and the actual scale on the page.

VECTOR SUBTRACTION:

then vector –A is a vector of the same magnitude as A
but opposite direction.

In order to find the difference between two vectors, add
the negative of the second vector to the first.

EXAMPLE
4:A Maserati (car) is moving due East at 20 ms-1.A short time later it is moving due North at 20 ms-1.Calculate the change in velocity of the Maserati.

To find the change in size of any quantity, you subtract
the initial size of the quantity from the final size.Obviously, with vector quantities you must do a vector
subtraction not just an arithmetic one, since vectors possess both size and
direction.

Change in velocity = final
velocity – initial velocity

This should really be written as:

Change in velocity = final
velocity + (– initial velocity)

since that is how we draw the vector diagram.We literally add the negative of the initial velocity to the final
velocity.Study the vector diagram
below to ensure you understand the process of vector subtraction.

Using Pythagoras’ Theorem and basic trigonometry as shown
in Example 1 above, we find that the change in velocity of the Maserati is
28.3 ms-1 at an angle of 45o West of North.

Note that even though the car has the same initial and
final speeds, because the direction of the car has changed, so too has its
velocity.

VECTOR COMPONENTS:

Sometimes
we are only interested in part of a vector rather than all of it.For instance, if we push a car that has run out of petrol, we apply a
force to the car.However, if we
are not careful some of the force we apply pushes down vertically on the car and
the rest of it pushes horizontally on the car.Obviously, we are trying to maximise the component of the force that is
applied horizontally.The angle at which we apply force to the car will determine
how much of our force is applied horizontally.

Any
vector may be broken into two component vectors at right angles to each other.These components are called the rectangular components of the vector.The rectangular components of a vector add up to the original vector.

Consider,
the example we used above.We may
push on the car with a force
F
at an angle of q
to the horizontal, as shown below.

The force F may be broken
into its vertical and horizontal components as shown below.

The magnitude of each component can then be calculated
using simple trigonometry.The size
of the vertical component of F is Fsinq.The size of the horizontal component of F,
the one that must overcome the force of friction if we are to move the car
forward, is Fcosq.

EXAMPLE 5:
A block of mass 20 kg is being pulled up an inclined plane by a rope inclined at
30o to the plane’s surface as shown in the diagram below.

The plane is inclined at 45o to the
horizontal.The friction force, F,
opposing the block’s motion is 10 N.Determine
the tension, T, in the rope if the net acceleration, a,
of the block up the plane is 4 ms-2.

You will observe that we have resolved two vectors in the diagram into
rectangular components – the tension, T, in the rope and
the weight, W, of the block.The rectangular components we have chosen are those acting parallel to
and perpendicular to the inclined plane.These
components are the most useful ones in a situation like this.

Now the total force acting down the plane is the
sum of the friction force, F, and the component of the
weight force of the block acting down the plane (W
sinq).So, from the diagram we have:

FD
= F + mg sinq
(since W = mg)

FD
= 10 + 20 x 9.8 x sin 45o

FD
= 148.59 N

Total force acting up the plane must be:

FU =
ma +FD

FU = (20 x 4) + 148.59

FU = 228.59 N

Note that the logic we have used to obtain an expression
for FUis
as follows.The force up the plane MUST
be sufficient to overcome the 148.59 N force down the plane and to provide the
required force of 80 N to give the block the correct acceleration up the plane.

Now we are in a position to calculate the tension in
the rope.The total force up
the plane FUis actually the component of the tension force parallel to the plane.
This is the part of the tension force that is applied parallel to the plane.

You may also like to check out this
applet,
which helps you to understand vector addition & subtraction. When you
get to the site, click on the "Applet Menu" button at the
top left of the page. The menu items will appear. Select "Some
Math", "Vectors", "Vector Addition". Read the
instructions & run the applet.

Tensions In Strings (Extension
Topic)

The
following section is not essential to the Syllabus. It is however a very
good section to do with students if time permits. It gives students an
opportunity to expand their understanding of both the usefulness of vectors in
Physics and how to analyze objects under the influence of gravity in different
situations.

Consider a mass, m,
supported by a thin, inextensible string of negligible mass.The two forces acting on the mass are T,
the tension in the string acting upwards and mg,
the weight force acting downwards on the mass.

The vector equation describing the net
force acting on the mass is best studied in three separate cases.

Case 1:
The mass is supported by the string and there is no acceleration.The mass could be at rest or could be moving up or down at constant
velocity.In each case the net force
equation is the same.The tension
upwards is exactly balancing the weight force downwards.

Case 2: The
mass is accelerating downwards with net acceleration, a.The tension upwards in the string is not sufficient to fully balance the
weight force downwards.

Case 3:
The mass is accelerating upwards with net acceleration, a.Here, the tension upwards in the string is doing two jobs.It is fully balancing the weight force downwards and supplying the
required force upwards to accelerate the mass with the net acceleration of a.

Many problems in Physics can be
solved by applying the knowledge summarized above.Consider the following two examples.

EXAMPLE
1: Two masses X of 10kg and Y of24kg are connected by a light inextensible string.

X and Y hang on opposite sides of a
frictionless pulley as shown above. Determine:
(a) the
net acceleration of the system of masses
(b)the
magnitude of the tension, T, in the string.

Assume that the
acceleration due to gravity is 9.8 ms-2.

SOLUTION:
There are two slightly different approaches possible.

Solution
1:The
Intuitive Approach

Firstly,
determine the net force on the system, the total mass of the system and then
obtain the net acceleration from F
= ma.Once
the net acceleration of the system is known the tension in the string can be
found by realising that the tension upwards on the left side of the pulley must
balance the weight force down on the mass & supply sufficient force to
accelerate the mass upwards with the net acceleration of the system.

(a) Force of gravity on 10kg mass = 10 x
9.8 = 98N down on left side of pulley.

Force of gravity on 24kg mass = 24 x 9.8 =
235.2N down on right side of pulley.

Thus, the net force, F, applied to the
system of two masses by gravity:

F
= 235.2 – 98=137.2 N down on right side of pulley.

Total
mass of system upon which this net force acts = 10 + 24 = 34kg.

Therefore
from F = ma, the net acceleration of the system of two masses is:

a = F / m= 137.2 / 34=4.035 ms-2.

Note that as
you get used to using this method, it really only takes a couple of lines of
working at the most.

(b)
Once the acceleration is known the tension can be calculated from either mass.Let’s use the 10kg mass first.Clearly,
the tension in the string on this side of the pulley must be sufficient to
balance the acceleration due to gravity down on the 10kg mass and to accelerate
the 10kg mass upwards at 4.035 ms-2.Therefore,

Tension, T = (10 x 9.8) + (10 x 4.035) = 138.35 N upwards.

OR
if we decided to use the 24kg mass instead - the tension in the string on the
right side of the pulley must be sufficient to balance the acceleration due to
gravity down on the 24kg mass less the 4.035 ms-2 that the mass is being permitted to accelerate downwards already.Therefore,

Tension, T = (24 x 9.8) – (24 x 4.035) = 138.36 N upwards.

This
is the same (to one decimal place) as the answer we obtained using the other
mass.This must be the case.It does not matter which mass you use to calculate the tension, you must
get the same answer in both cases because there is only one string and can
therefore be only one tension.

The
tension in the string is therefore 138.4 N
to one decimal place.Any
discrepancy in the above answers after the first decimal place is simply due to
rounding off the (137.2 / 34) calculation for the net acceleration in the first
place.

Again I
stress that the whole solution (parts a & b) I have demonstrated here would
normally take no more than four to five lines of calculation.It is the explanation I have made during the solution that has greatly
increased the space used.

Solution
2:The Mathematical Approach

(a)
First we write down the two vector equations of motion for the masses.

For
mass X:T
– mxg = mxa-
(1)

For
mass Y:myg
– T = mya- (2)

Now,
we solve these equations simultaneously.So,
adding equation 1 and 2 together we have:

a=(myg– mxg) / (mx+ my
) = (24 x 9.8 – 10 x 9.8) / (10 + 24)

= 4.035 ms-2

(b)
Then from either
equation 1 or 2, we can calculate the value of T.

From
equation 1:T = (10 x 9.8) + (10 x 4.035) = 138.35 N

This
second solution is probably the more mathematically pleasing to the eye.For my liking though, the previous solution is the more physically
intuitive method.Both solutions are
equally acceptable and in the end it’s really only the setting out that
differs.Suit yourself as to which
one you use.You will find the more
mathematical approach is safer as the problems become more complex.

EXAMPLE
2: Three masses of 2kg, 4kg and 6kg
are connected by three light inextensible strings, X, Y and Z as shown below.The masses are supported from the roof of a lift of mass 1000kg.The lift is accelerating downwards with a net acceleration of 3 ms-2.

Determine:

a)The tension in string X.

b)The tension in string Y.

c)The tension in string Z.

d)The tension in the supporting cable.

SOLUTION:

Note that since the lift is accelerating
downwards at 3 ms-2, we can write for any string
supporting a mass inside the lift or indeed for the supporting cable itself
that:

Circular Motion

An object moving in
a circular path at constant speed is said to be executing uniform circular
motion.Obviously, although the
speed is constant, the velocity is not, since the direction of the motion is
always changing.It can be shown
that for an object executing uniform circular motion (UCM), the acceleration
keeping the object in its circular path is given by:

ac
= v2/r

where ac is
called the centripetal ("centre-seeking") acceleration, v =
speed of the object and r = radius of the circular path.As the name implies, centripetal acceleration is directed towards the
centre of the circle.

Using the fact that force can be
written as F = ma, the centripetal force, Fc,
acting on an object undergoing UCM is given by:

where m = mass of the object.This force is also directed
towards the centre of the circle.

Example: A
car of mass 900 kg moves at a constant speed of 25 m/s around a circular curve
of radius 50 m. Calculate the centripetal force acting on the car.
(11250 N, towards the centre of the circle)

ENERGY

Energy and work are closely related quantities.An object can do work only if it has energy.Energy, then, is the property of a system that is a measure of its
capacity for doing work.The amount
of energy an object has is equal to the amount of work it can do.Like work,energy is a scalar quantity with an SI Unit of the
joule (J).

Energy has several forms:
electric energy, chemical energy, heat energy, nuclear energy, radiant energy (ie
EM radiation such as light), mechanical energy (eg kinetic energy) and sound
energy (ie the kinetic energy of the vibration of the air).In a closed system (ie one in which no mass enters or leaves), energy
can neither be created nor destroyed, although it may be transformed from one
form into another.This is called
the Law of Conservation of Energy.

KINETIC ENERGY

Kinetic energy is the name
given to the energy associated with a moving body.It can be shown that the amount of kinetic energy possessed
by a body is given by:

where m = mass of the body and v =
velocity the body.

Moving vehicles have kinetic
energy.Consider a small car of
mass 920 kg moving at a constant speed of 60 km/h (16.67 m/s).The kinetic energy of the car can be calculated as:

Ek
= 0.5 x 920
x 16.672

Ek
= 1.28 x 105
J

To change the velocity of a
moving body, or to set a body at rest into motion, a net force must be applied to it and work
must be done on it.The work done
on the body is
equivalent to the change in kinetic energy of the body.

POTENTIAL ENERGY

Stored energy is called potential
energy, since it has the potential to do work for us.Examples of potential energy include: the energy stored in a stretched
(or compressed) spring; the chemical energy stored in a car battery; the energy
stored in the water in a damn above a hydroelectric power station; and the
energy stored in the chemical bonds holding compounds together.

ENERGY
TRANSFORMATIONS

Energy transformations (changes)
are an important aspect in understanding motion.In a car battery, chemical potential energy is transformed
into electrical energy.In an
internal combustion engine (such as a car engine) the chemical potential energy
stored in petrol is transformed among other things into mechanical energy in the
form of kinetic energy of motion.

When cars collide, various energy
transformations take place.Some of
the kinetic energy (KE) of the cars is transformed into sound – we hear the
collision.Some KE is changed into
radiant energy (light energy) – we see sparks fly as the wreckage scrapes
along the ground.Some of the KE is
transformed into heat – the friction produced by metal scraping on metal and
tyres gripping the road under heavy braking produces heat.Some KE is transformed into potential energy of deformation – the car
bodies are compressed, compacted and twisted during the collision and some of
the KE is stored in the deformed wreckage.In a worst-case scenario, where an explosion takes place, chemical energy
stored in the fuel is converted into kinetic energy (and sound, heat, light) as
parts of the wreckage are flung far and wide.

MOMENTUM:

Everyday experience tells us that
both the mass and velocity of an object are important in
determining things like (i) how hard it is to stop the object or (ii) the effect
the object has in a collision with another object.An 85 kg man running at 5 m/s is a lot harder to stop than a 15 kg six
year old child running at the same speed.A
50 gram bullet fired from a rifle with a muzzle velocity of 500 m/s will do a
lot more damage than an identical bullet thrown at the target by hand.

Isaac Newton spoke of the “quantity
of motion” of an object.Today
we define the momentum of an object to be the product of mass (m) and velocity
(v).

Momentum is a vector
quantity with SI Units of kgms-1 (or Ns, since 1N = 1kgms-2).

Newton’s 2nd Law can
be re-written as:

where Dp = the change in momentum of the object and Dt
= the time taken for the change in momentum to occur.

This quantity Dp
(the change in momentum) is given the name impulse.Clearly, from the above equation, impulse, I, is defined as the product
of force and time and has SI Units of Ns.Impulse is a vector quantity.

CONSERVATION
OF MOMENTUM

According to Newton’s 1st
and 2nd Laws of motion, there is no change in momentum without
the action of a net external force.Thus,
if no net external force acts on a particular system, the total momentum of the
system must be constant.This
is known as the principle of the conservation of linear momentum.

A system on which the net external
force is zero is given a special name.Such
a system is called an isolated system. So, another way to express the
principle of the conservation of linear momentum is to say that within an
isolated system the total momentum is a constant.This principle is applicable to many important physical situations.

CONSERVATION
OF MOMENTUM DURING COLLISIONS

One important physical situation
to which the principle of the conservation of linear momentum is applicable is
the case of collisions between bodies.In
such cases, if we assume that no external net force acts during the collision,
we can say that the total momentum of the system before collision equals the
total momentum of the system after collision.This proves to be an extremely useful starting point for
analysing many collision situations.

To see that momentum is
conserved during collisions we can use Newton’s 3rd Law.Consider a collision between two particles, A and B,
as shown below.

During the brief collision these
particles exert large forces on one another.At any instant FAB
is the force exerted on A by B and FBA is the force exerted on B
by A.By
Newton’s 3rd Law these forces at any instant are equal in magnitude
but opposite in direction.

The change in momentum of A
resulting from the collision is:

in which the bar above the FAB
indicates that we are taking the average value ofFAB
during the time interval of the collision, Dt.

The change in momentum of B
resulting from the collision is:

in which the bar above the FBA
indicates that we are taking the average value of FBA
during the time interval of the collision, Dt.

Note that it is necessary to take
the average value of the collision forces since the magnitudes of both
forces will vary over the duration of the collision.

If no other forces act on the
particles, then DpA and DpB
give the total change in momentum for each particle.But we have seen that at each instant:

FAB =
- FBA

So
that:

And therefore that:DpA
= - DpB
.

If we consider the two particles
as an isolated system, the total momentum of the system is:

P = pA + pB

And the total change in momentum
of the system as a result of the collision is zero, that is:

DP=DpA
+DpB = 0.

Thus,
using Newton’s 3rd Law and our knowledge of impulse
we have shown that if there are no external forces, the total momentum of the
system is not changed by the collision.Therefore, as we said before, if we assume that no external net force
acts during the collision, we can say that the total momentum of the system
before collision equals the total momentum of the system after collision.

How accurate is it though to
assume that no external net force acts on a system during a collision?When a golf club strikes a golf ball surely there are external forces
that act on the system of club + ball?Indeed
there are: gravity and friction are two obvious forces that act on
both club and ball.So how can we
simply ignore these forces?

The answer is that it is safe
to neglect these external forces during the collision and to assume that
momentum is conserved provided, as is almost always the case, that the external
forces are negligible compared to the impulsive forces of collision.If the external forces are negligible compared to the impulsive forces,
then the change in momentum of a particle during a collision arising from an
external force is negligible compared to the change in momentum of that particle
arising from the impulsive force of collision.

In the case of the golf club
striking the golf ball, the collision lasts only a tiny fraction of a second.Since the observed change in momentum is large and the time of collision
is small, it follows from the impulse equation:

Dp
= F Dt

that the average impulsive force F
is relatively large.Compared to
this force, the external forces of gravity and friction are negligible.During the collision we can safely ignore these external forces in
determining the change in motion of the ball; the shorter the collision time,
the more accurate this assumption becomes.

In practice, we can apply the
principle of momentum conservation during collisions if the time of collision is
small enough.

*NOTE:
This section of notes on Conservation of Momentum During Collisions has followed
the treatment on pages 213-214 in “Physics Parts I & II Combined” by D
Halliday & R Resnick (Wiley, 1966).

INERTIA

As we have seen, Newton’s 1st
Law states that an object in uniform motion will remain in uniform motion and an
object at rest will remain at rest, unless acted upon by an external, net force.This ability of a body to resist changes in its state of motion is
called the inertia of the body.The
inertia of a car, for instance, is its tendency to remain in uniform motion or remain at
rest.The fact that bodies possess inertia has important
consequences when dealing with moving vehicles.

A moving vehicle is a complex
body.It consists of the vehicle
body itself, the driver (and passengers) and other objects carried in the car.If the driver or passengers or other objects are not restrained, they
will continue to move at whatever speed the car is travelling, even after the
application of the brakes.Let us consider some questions:

uWhat are some of the dangers presented by loose objects in
vehicles?

uMost people, when they see an unrestrained object fly off the car
seat when they slam on the brakes, would say that it got pushed off the seat (ie
some sort of force acted on the object to move it off the seat).Is this an accurate account of the physics of the situation?Explain in terms of Newton’s 1st Law.

uWhy do you think Newton’s 1st Law of Motion is not
applied correctly in many real world situations (like the one mentioned above)?

uWhat is the function of an inertia reel safety belt?Where could we obtain information on how they work and their
effectiveness?

uDescribe a possible experiment we could do to assess the relative
effectiveness of lap, lap-sash and harness seatbelts in reducing the effects of
inertia in a collision.

uList the features of a modern car that are designed to reduce or
avoid the effects of a collision.Where
would we obtain information to use in assessing the relative effectiveness of
these features?

uAssess the reasons for the introduction of low speed zones in built-up
areas and the addition of air bags and crumple zones to vehicles with reference
to the concepts of impulse and momentum.

See
http://www.science.org.au/nova/058/058key.htm
for a very interesting article on the physics of speeding cars based on work
done by a statistician. You may like to compare this article with this one
at http://www.ibiblio.org/rdu/speedsci.html
which was written by a theoretical physicist. The second article uses more
appropriate equations than the first and is probably more accurate in its
estimates of risk. Interesting, isn't it? Please realize that this
second article is not advocating speeding for the sake of speeding. It is
attempting to provide realistic estimates of the risks associated with
speeding. It presents one possible analysis. There are others, as
the article itself points out.

In
my opinion, when on the roads, stick to the speed limits, obey all the
road rules & concentrate on what you are doing. This applies no matter
how old you are or how experienced a driver you happen to be BUT especially if
you are a young person, who has just started driving!!!